Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int nmax = 503;
long long a[nmax][nmax];
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string& s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
void debug_out() { cout << '\n'; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cout << " " << to_string(H);
  debug_out(T...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, x, t, m, y, z, k;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  auto p = [&](vector<array<long long, 3>> foo) {
    for (auto it : foo) {
      for (int i = 0; i < it[2]; i++) {
        if (it[0] == 0) {
          cout << "row ";
        } else {
          cout << "col ";
        }
        cout << it[1] + 1 << " ";
        cout << '\n';
      }
    }
  };
  vector<array<long long, 3>> best;
  long long ans = (1ll << 60) - 1;
  for (int k = 0; k < nmax; k++) {
    long long curr = 0;
    vector<array<long long, 3>> curr_moves;
    vector<long long> r(n, -1), c(m, -1);
    curr += k;
    r[0] = k;
    curr_moves.push_back({0, 0, k});
    bool ok = true;
    for (int i = 0; (i < n) && ok; i++) {
      for (int j = 0; (j < m) && ok; j++) {
        if (r[i] != -1 && c[j] != -1 && r[i] + c[j] != a[i][j]) {
          ok = false;
          break;
        } else if (r[i] != -1 && c[j] != -1)
          continue;
        if (r[i] != -1) {
          c[j] = a[i][j] - r[i];
          curr += c[j];
          curr_moves.push_back({1, j, c[j]});
          if (c[j] < 0) {
            ok = false;
            break;
          }
        } else if (c[j] != -1) {
          r[i] = a[i][j] - c[j];
          curr += r[i];
          curr_moves.push_back({0, i, r[i]});
          if (r[i] < 0) {
            ok = false;
            break;
          }
        } else {
          ok = false;
          break;
        }
      }
    }
    if (ok) {
      if (curr < ans) {
        ans = curr;
        best = curr_moves;
      }
    }
  }
  if (ans == (1ll << 60) - 1) {
    cout << -1;
    return 0;
  }
  cout << ans << '\n';
  p(best);
  return 0;
}